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User: Gabin Kolly

Gabin Kolly has authored 1 sequences.

A322522 a(n) is the minimal absolute difference between n and each of the powers of the previous terms; a(1) = 1.

1, 1, 2, 0, 1, 2, 1, 0, 1, 2, 3, 3, 3, 2, 1, 0, 1, 2, 3, 4, 5, 3, 2, 1, 0, 1, 0, 1, 2, 2, 1, 0, 1, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1

Offset: 1

Gabin Kolly, Aug 28 2019

a(n) <= ceiling(sqrt(n)); if n is between k^2 and (k+1)^2, we have min(n - k^2, (k+1)^2 - n) <= k < ceiling(sqrt(n)).
Using the fact that the density of the nontrivial powers over the integers is 0, and that the density of cubes and higher powers among the nontrivial powers is 0, we can show that there are an infinite number of integers i such that i is not a nontrivial power, and there is no cube or higher power between (i-1)^2 and i^2. We then have a((i-1)^2 + i) = i = ceiling(sqrt((i-1)^2 + i)). Therefore there are infinitely many numbers n such that a(n) = ceiling(sqrt(n)).
a(n) - a(n-1) <= 1.
For the first 10000 terms, indices for a(n) = 1 correspond to 85 of 90 values of A227802(m). - Bill McEachen, Feb 26 2024

			For n = 4, we have a(4) = 0, because a(3) = 2, and 2^2 - 4 = 0.
For n = 6, we have a(6) = 0, because there are only 0, 1 and 2 in the first 5 terms, and therefore the closest power is 2^2 = 4 or 2^3 = 8, with an absolute difference of 2.

Gabin Kolly, Table of n, a(n) for n = 1..10000

Cf. A301573 (distance from n to the nearest nontrivial power).

comparePowers[n_, m_] :=
If[n <= 1, m - n, a = n; While[a < m, a *= n];
  Min[m - a/n, a - m]]; list = {1}; cleanList = {1}; Do[
list = Append[list,
   Min[comparePowers[#, Length[list] + 1] & /@ cleanList]];
If[Last[list] > Last[cleanList],
  cleanList = Append[cleanList, Last[list]]], 9999]; Print[list]

Let b(n) be the first time that n appears in the sequence; then b(n) ~ n^2.

cp's OEIS Frontend

User: Gabin Kolly

A322522 a(n) is the minimal absolute difference between n and each of the powers of the previous terms; a(1) = 1.

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